Question Number 89991 by jagoll last updated on 20/Apr/20
$$\mathrm{if}\:\mathrm{a},\mathrm{b}\:>\:\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{find}\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{ab}} \\ $$
Commented by john santu last updated on 20/Apr/20
$${ab}\:=\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\:,\:{since}\:{a},{b}\:>\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{3}}{b}^{\mathrm{2}} \:\Rightarrow{ab}\:=\:\sqrt{\frac{\mathrm{5}{b}^{\mathrm{4}} }{\mathrm{3}}}\:=\:\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{15}}}{\mathrm{3}} \\ $$$${then}\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}\:=\:\frac{\frac{\mathrm{8}{b}^{\mathrm{2}} }{\mathrm{3}}}{\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{15}}}{\mathrm{3}}} \\ $$$$=\:\frac{\mathrm{8}}{\:\sqrt{\mathrm{15}}} \\ $$
Commented by mr W last updated on 20/Apr/20
$$\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{a}}{{b}}=\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}=\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{15}}}=\frac{\mathrm{8}\sqrt{\mathrm{15}}}{\mathrm{15}} \\ $$